3.1.42 \(\int \frac {(e x)^m (A+B x^2)}{(a+b x^2)^2 (c+d x^2)^3} \, dx\) [42]

Optimal. Leaf size=452 \[ \frac {d (2 A b c-3 a B c+a A d) (e x)^{1+m}}{4 a c (b c-a d)^2 e \left (c+d x^2\right )^2}+\frac {(A b-a B) (e x)^{1+m}}{2 a (b c-a d) e \left (a+b x^2\right ) \left (c+d x^2\right )^2}+\frac {d \left (A \left (4 b^2 c^2-a^2 d^2 (3-m)+a b c d (11-m)\right )-a B c (b c (11-m)+a d (1+m))\right ) (e x)^{1+m}}{8 a c^2 (b c-a d)^3 e \left (c+d x^2\right )}+\frac {b^2 (A b (b c (1-m)-a d (7-m))+a B (a d (5-m)+b c (1+m))) (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )}{2 a^2 (b c-a d)^4 e (1+m)}-\frac {d \left (b^2 c^2 (B c (3-m)-A d (7-m)) (5-m)-a^2 d^2 (1-m) (A d (3-m)+B c (1+m))+2 a b c d \left (B c \left (5+4 m-m^2\right )+A d \left (7-8 m+m^2\right )\right )\right ) (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )}{8 c^3 (b c-a d)^4 e (1+m)} \]

[Out]

1/4*d*(A*a*d+2*A*b*c-3*B*a*c)*(e*x)^(1+m)/a/c/(-a*d+b*c)^2/e/(d*x^2+c)^2+1/2*(A*b-B*a)*(e*x)^(1+m)/a/(-a*d+b*c
)/e/(b*x^2+a)/(d*x^2+c)^2+1/8*d*(A*(4*b^2*c^2-a^2*d^2*(3-m)+a*b*c*d*(11-m))-a*B*c*(b*c*(11-m)+a*d*(1+m)))*(e*x
)^(1+m)/a/c^2/(-a*d+b*c)^3/e/(d*x^2+c)+1/2*b^2*(A*b*(b*c*(1-m)-a*d*(7-m))+a*B*(a*d*(5-m)+b*c*(1+m)))*(e*x)^(1+
m)*hypergeom([1, 1/2+1/2*m],[3/2+1/2*m],-b*x^2/a)/a^2/(-a*d+b*c)^4/e/(1+m)-1/8*d*(b^2*c^2*(B*c*(3-m)-A*d*(7-m)
)*(5-m)-a^2*d^2*(1-m)*(A*d*(3-m)+B*c*(1+m))+2*a*b*c*d*(B*c*(-m^2+4*m+5)+A*d*(m^2-8*m+7)))*(e*x)^(1+m)*hypergeo
m([1, 1/2+1/2*m],[3/2+1/2*m],-d*x^2/c)/c^3/(-a*d+b*c)^4/e/(1+m)

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Rubi [A]
time = 0.89, antiderivative size = 452, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {593, 598, 371} \begin {gather*} \frac {d (e x)^{m+1} \left (A \left (-a^2 d^2 (3-m)+a b c d (11-m)+4 b^2 c^2\right )-a B c (a d (m+1)+b c (11-m))\right )}{8 a c^2 e \left (c+d x^2\right ) (b c-a d)^3}-\frac {d (e x)^{m+1} \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {d x^2}{c}\right ) \left (-a^2 d^2 (1-m) (A d (3-m)+B c (m+1))+2 a b c d \left (A d \left (m^2-8 m+7\right )+B c \left (-m^2+4 m+5\right )\right )+b^2 c^2 (5-m) (B c (3-m)-A d (7-m))\right )}{8 c^3 e (m+1) (b c-a d)^4}+\frac {b^2 (e x)^{m+1} \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {b x^2}{a}\right ) (A b (b c (1-m)-a d (7-m))+a B (a d (5-m)+b c (m+1)))}{2 a^2 e (m+1) (b c-a d)^4}+\frac {d (e x)^{m+1} (a A d-3 a B c+2 A b c)}{4 a c e \left (c+d x^2\right )^2 (b c-a d)^2}+\frac {(e x)^{m+1} (A b-a B)}{2 a e \left (a+b x^2\right ) \left (c+d x^2\right )^2 (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((e*x)^m*(A + B*x^2))/((a + b*x^2)^2*(c + d*x^2)^3),x]

[Out]

(d*(2*A*b*c - 3*a*B*c + a*A*d)*(e*x)^(1 + m))/(4*a*c*(b*c - a*d)^2*e*(c + d*x^2)^2) + ((A*b - a*B)*(e*x)^(1 +
m))/(2*a*(b*c - a*d)*e*(a + b*x^2)*(c + d*x^2)^2) + (d*(A*(4*b^2*c^2 - a^2*d^2*(3 - m) + a*b*c*d*(11 - m)) - a
*B*c*(b*c*(11 - m) + a*d*(1 + m)))*(e*x)^(1 + m))/(8*a*c^2*(b*c - a*d)^3*e*(c + d*x^2)) + (b^2*(A*b*(b*c*(1 -
m) - a*d*(7 - m)) + a*B*(a*d*(5 - m) + b*c*(1 + m)))*(e*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2,
-((b*x^2)/a)])/(2*a^2*(b*c - a*d)^4*e*(1 + m)) - (d*(b^2*c^2*(B*c*(3 - m) - A*d*(7 - m))*(5 - m) - a^2*d^2*(1
- m)*(A*d*(3 - m) + B*c*(1 + m)) + 2*a*b*c*d*(B*c*(5 + 4*m - m^2) + A*d*(7 - 8*m + m^2)))*(e*x)^(1 + m)*Hyperg
eometric2F1[1, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)])/(8*c^3*(b*c - a*d)^4*e*(1 + m))

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 593

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_)*((e_) + (f_.)*(x_)^(n_)), x
_Symbol] :> Simp[(-(b*e - a*f))*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*g*n*(b*c - a*d)*(p +
1))), x] + Dist[1/(a*n*(b*c - a*d)*(p + 1)), Int[(g*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f)
*(m + 1) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c,
d, e, f, g, m, q}, x] && IGtQ[n, 0] && LtQ[p, -1]

Rule 598

Int[(((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n_)))/((c_) + (d_.)*(x_)^(n_)), x_Sy
mbol] :> Int[ExpandIntegrand[(g*x)^m*(a + b*x^n)^p*((e + f*x^n)/(c + d*x^n)), x], x] /; FreeQ[{a, b, c, d, e,
f, g, m, p}, x] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {(e x)^m \left (A+B x^2\right )}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^3} \, dx &=\frac {(A b-a B) (e x)^{1+m}}{2 a (b c-a d) e \left (a+b x^2\right ) \left (c+d x^2\right )^2}-\frac {\int \frac {(e x)^m \left (2 a A d-A b c (1-m)-a B c (1+m)-(A b-a B) d (5-m) x^2\right )}{\left (a+b x^2\right ) \left (c+d x^2\right )^3} \, dx}{2 a (b c-a d)}\\ &=\frac {d (2 A b c-3 a B c+a A d) (e x)^{1+m}}{4 a c (b c-a d)^2 e \left (c+d x^2\right )^2}+\frac {(A b-a B) (e x)^{1+m}}{2 a (b c-a d) e \left (a+b x^2\right ) \left (c+d x^2\right )^2}-\frac {\int \frac {(e x)^m \left (2 \left (A \left (8 a b c d-2 b^2 c^2 (1-m)-a^2 d^2 (3-m)\right )-a B c (2 b c+a d) (1+m)\right )-2 b d (2 A b c-3 a B c+a A d) (3-m) x^2\right )}{\left (a+b x^2\right ) \left (c+d x^2\right )^2} \, dx}{8 a c (b c-a d)^2}\\ &=\frac {d (2 A b c-3 a B c+a A d) (e x)^{1+m}}{4 a c (b c-a d)^2 e \left (c+d x^2\right )^2}+\frac {(A b-a B) (e x)^{1+m}}{2 a (b c-a d) e \left (a+b x^2\right ) \left (c+d x^2\right )^2}+\frac {d \left (A \left (4 b^2 c^2-a^2 d^2 (3-m)+a b c d (11-m)\right )-a B c (b c (11-m)+a d (1+m))\right ) (e x)^{1+m}}{8 a c^2 (b c-a d)^3 e \left (c+d x^2\right )}-\frac {\int \frac {(e x)^m \left (-2 \left (a B c \left (4 b^2 c^2-a^2 d^2 (1-m)+a b c d (9-m)\right ) (1+m)-A \left (24 a b^2 c^2 d-4 b^3 c^3 (1-m)-a^2 b c d^2 \left (11-12 m+m^2\right )+a^3 d^3 \left (3-4 m+m^2\right )\right )\right )-2 b d (1-m) \left (A \left (4 b^2 c^2-a^2 d^2 (3-m)+a b c d (11-m)\right )-a B c (b c (11-m)+a d (1+m))\right ) x^2\right )}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx}{16 a c^2 (b c-a d)^3}\\ &=\frac {d (2 A b c-3 a B c+a A d) (e x)^{1+m}}{4 a c (b c-a d)^2 e \left (c+d x^2\right )^2}+\frac {(A b-a B) (e x)^{1+m}}{2 a (b c-a d) e \left (a+b x^2\right ) \left (c+d x^2\right )^2}+\frac {d \left (A \left (4 b^2 c^2-a^2 d^2 (3-m)+a b c d (11-m)\right )-a B c (b c (11-m)+a d (1+m))\right ) (e x)^{1+m}}{8 a c^2 (b c-a d)^3 e \left (c+d x^2\right )}-\frac {\int \left (\frac {8 b^2 c^2 (-A b (b c (1-m)-a d (7-m))-a B (a d (5-m)+b c (1+m))) (e x)^m}{(b c-a d) \left (a+b x^2\right )}+\frac {2 a d \left (b^2 c^2 (B c (3-m)-A d (7-m)) (5-m)-a^2 d^2 (1-m) (A d (3-m)+B c (1+m))+2 a b c d \left (B c \left (5+4 m-m^2\right )+A d \left (7-8 m+m^2\right )\right )\right ) (e x)^m}{(b c-a d) \left (c+d x^2\right )}\right ) \, dx}{16 a c^2 (b c-a d)^3}\\ &=\frac {d (2 A b c-3 a B c+a A d) (e x)^{1+m}}{4 a c (b c-a d)^2 e \left (c+d x^2\right )^2}+\frac {(A b-a B) (e x)^{1+m}}{2 a (b c-a d) e \left (a+b x^2\right ) \left (c+d x^2\right )^2}+\frac {d \left (A \left (4 b^2 c^2-a^2 d^2 (3-m)+a b c d (11-m)\right )-a B c (b c (11-m)+a d (1+m))\right ) (e x)^{1+m}}{8 a c^2 (b c-a d)^3 e \left (c+d x^2\right )}+\frac {\left (b^2 (A b (b c (1-m)-a d (7-m))+a B (a d (5-m)+b c (1+m)))\right ) \int \frac {(e x)^m}{a+b x^2} \, dx}{2 a (b c-a d)^4}-\frac {\left (d \left (b^2 c^2 (B c (3-m)-A d (7-m)) (5-m)-a^2 d^2 (1-m) (A d (3-m)+B c (1+m))+2 a b c d \left (B c \left (5+4 m-m^2\right )+A d \left (7-8 m+m^2\right )\right )\right )\right ) \int \frac {(e x)^m}{c+d x^2} \, dx}{8 c^2 (b c-a d)^4}\\ &=\frac {d (2 A b c-3 a B c+a A d) (e x)^{1+m}}{4 a c (b c-a d)^2 e \left (c+d x^2\right )^2}+\frac {(A b-a B) (e x)^{1+m}}{2 a (b c-a d) e \left (a+b x^2\right ) \left (c+d x^2\right )^2}+\frac {d \left (A \left (4 b^2 c^2-a^2 d^2 (3-m)+a b c d (11-m)\right )-a B c (b c (11-m)+a d (1+m))\right ) (e x)^{1+m}}{8 a c^2 (b c-a d)^3 e \left (c+d x^2\right )}+\frac {b^2 (A b (b c (1-m)-a d (7-m))+a B (a d (5-m)+b c (1+m))) (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )}{2 a^2 (b c-a d)^4 e (1+m)}-\frac {d \left (b^2 c^2 (B c (3-m)-A d (7-m)) (5-m)-a^2 d^2 (1-m) (A d (3-m)+B c (1+m))+2 a b c d \left (B c \left (5+4 m-m^2\right )+A d \left (7-8 m+m^2\right )\right )\right ) (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )}{8 c^3 (b c-a d)^4 e (1+m)}\\ \end {align*}

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Mathematica [A]
time = 1.64, size = 267, normalized size = 0.59 \begin {gather*} \frac {x (e x)^m \left (a b^2 c^3 (b B c-3 A b d+2 a B d) \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )-a^2 b c^2 d (b B c-3 A b d+2 a B d) \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )+(b c-a d) \left (b^2 (A b-a B) c^3 \, _2F_1\left (2,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )-a^2 d \left (c (b B c-2 A b d+a B d) \, _2F_1\left (2,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )+(b c-a d) (B c-A d) \, _2F_1\left (3,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )\right )\right )\right )}{a^2 c^3 (b c-a d)^4 (1+m)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((e*x)^m*(A + B*x^2))/((a + b*x^2)^2*(c + d*x^2)^3),x]

[Out]

(x*(e*x)^m*(a*b^2*c^3*(b*B*c - 3*A*b*d + 2*a*B*d)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)] - a
^2*b*c^2*d*(b*B*c - 3*A*b*d + 2*a*B*d)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)] + (b*c - a*d)*
(b^2*(A*b - a*B)*c^3*Hypergeometric2F1[2, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)] - a^2*d*(c*(b*B*c - 2*A*b*d + a*
B*d)*Hypergeometric2F1[2, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)] + (b*c - a*d)*(B*c - A*d)*Hypergeometric2F1[3, (
1 + m)/2, (3 + m)/2, -((d*x^2)/c)]))))/(a^2*c^3*(b*c - a*d)^4*(1 + m))

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {\left (e x \right )^{m} \left (B \,x^{2}+A \right )}{\left (b \,x^{2}+a \right )^{2} \left (d \,x^{2}+c \right )^{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^m*(B*x^2+A)/(b*x^2+a)^2/(d*x^2+c)^3,x)

[Out]

int((e*x)^m*(B*x^2+A)/(b*x^2+a)^2/(d*x^2+c)^3,x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*(B*x^2+A)/(b*x^2+a)^2/(d*x^2+c)^3,x, algorithm="maxima")

[Out]

integrate((B*x^2 + A)*(x*e)^m/((b*x^2 + a)^2*(d*x^2 + c)^3), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*(B*x^2+A)/(b*x^2+a)^2/(d*x^2+c)^3,x, algorithm="fricas")

[Out]

integral((B*x^2 + A)*(x*e)^m/(b^2*d^3*x^10 + (3*b^2*c*d^2 + 2*a*b*d^3)*x^8 + (3*b^2*c^2*d + 6*a*b*c*d^2 + a^2*
d^3)*x^6 + a^2*c^3 + (b^2*c^3 + 6*a*b*c^2*d + 3*a^2*c*d^2)*x^4 + (2*a*b*c^3 + 3*a^2*c^2*d)*x^2), x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)**m*(B*x**2+A)/(b*x**2+a)**2/(d*x**2+c)**3,x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*(B*x^2+A)/(b*x^2+a)^2/(d*x^2+c)^3,x, algorithm="giac")

[Out]

integrate((B*x^2 + A)*(x*e)^m/((b*x^2 + a)^2*(d*x^2 + c)^3), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (B\,x^2+A\right )\,{\left (e\,x\right )}^m}{{\left (b\,x^2+a\right )}^2\,{\left (d\,x^2+c\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^2)*(e*x)^m)/((a + b*x^2)^2*(c + d*x^2)^3),x)

[Out]

int(((A + B*x^2)*(e*x)^m)/((a + b*x^2)^2*(c + d*x^2)^3), x)

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